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liq [111]
3 years ago
10

Join my G i m k i t Code is 85963

Mathematics
1 answer:
Ludmilka [50]3 years ago
7 0

Answer

I joined but it said it was full

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(x+1)(2x-4)(1/x+1) = (x+1)(2x-4)(1-5/2x-4)
Sunny_sXe [5.5K]

Solution:

(x+1)(2x-4)(\frac{1}{x}+1)=(x+1)(2x-4)(1-\frac{5}{2x}-4)\\(x+1)(2x-4)(\frac{1}{x}+1)-(x+1)(2x-4)(1-\frac{5}{2x}-4) = 0\\(x+1)(2x-4)(\frac{1}{x}+1-(1-\frac{5}{2x}-4))=0\\(x+1)(2x-4)(\frac{1}{x}+1-(-3\frac{5}{2x}))=0\\(x+1)(2x-4)(\frac{1}{x}+1+3+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{1}{x}+4+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{5x^{2}+8x+2}{2x} )=0\\

x+1=0\\x=-1\\\\2x-4=0\\x=2\\\\\frac{5x^{2}+8x+2}{2x}=0\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}

Answer:

x=-1\\x=2\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}

<em>Hope this was helpful.</em>

8 0
3 years ago
Which one the answer?
kifflom [539]
D. Hope this helps ;)
5 0
3 years ago
Use lagrange multipliers to find the maximum volume of a rectangular box that is inscribed in a sphere of radius r
vesna_86 [32]

For the derivative tests method, assume that the sphere is centered at the origin, and consider the circular projection of the sphere onto the xy-plane. An inscribed rectangular box is uniquely determined

1

by the xy-coordinate of its corner in the first octant, so we can compute the z coordinate of this corner by

x2+y2+z2=r2 =⇒z= r2−(x2+y2). Then the volume of a box with this coordinate for the corner is given by

V = (2x)(2y)(2z) = 8xy r2 − (x2 + y2),

and we need only maximize this on the domain x2 + y2 ≤ r2. Notice that the volume is zero on the boundary of this domain, so we need only consider critical points contained inside the domain in order to carry this optimization out.

For the method of Lagrange multipliers, we optimize V(x,y,z) = 8xyz subject to the constraint x2 + y2 + z2 = r2<span>. </span>

7 0
3 years ago
Using a directrix of y = −2 and a focus of (1, 6), what quadratic function is created? Question 3 options:
deff fn [24]

If your directrix is a "y=" line, that means that the parabola opens either upwards or downwards (as opposed to the left or the right). Because it is in the character of a parabola to "hug" the focus, our parabola opens upwards. The vertex of a parabola sits exactly halfway between the directrix and the focus. Since our directrix is at y = -2 and the focus is at (1, 6) AND the parabola opens upward, the vertex is going to sit on the main transversal, which is also the "line" the focus sits on. The focus is on the line x = 1, so the vertex will also have that x coordinate. Halfway between the y points of the directrix and the focus, -2 and 6, respectively, is the y value of 2. So the vertex sits at (1, 2). The formula for this type of parabola is (x-h)^2=4p(y-k) where h and k are the coordinates of the vertex and p is the DISTANCE that the focus is from the vertex. Our focus is 4 units from the vertex, so p = 4. Filling in our h, k, and p: (x-1)^2=4(4)(y-2). Simplifying a bit gives us (x-1)^2=16(y-2). We can begin to isolate the y by dividing both sides by 16 to get \frac{1}{16}(x-1)^2=y-2. Then we can add 2 to both sides to get the final equation f(x)=\frac{1}{16}(x-1)^2+2, choice 4 from above.

7 0
3 years ago
Please write the answer​
Travka [436]

\sf \dfrac{3^x - 5 \times 3^{(x-2)}}{3^{(x-3)}} \\ \\ \longrightarrow \sf \dfrac{ {3}^{x} }{ {3}^{(x - 3)}} - \frac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)}  \\ \\ \longrightarrow \sf {3}^{[x - (x - 3)]} - \dfrac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ {3}^{(x - 3)} } \times \dfrac{ {3}^{x}}{9} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ \bigg(\dfrac{ {3}^{x} }{ {3}^{3} }\bigg) } \times \dfrac{ {3}^{x}}{9}\\ \\ \longrightarrow \sf {3}^{3} - \dfrac{ ({3}^{3})( 5)}{ {3}^{x} } \times \dfrac{ {3}^{x}}{9}\\ \\ \sf \longrightarrow {3}^{3} - (5 \times 3) \\  \\ \longrightarrow \sf \: 27 - 15 \\  \\ \longrightarrow \leadsto{\underline{\boxed{\sf{ \pink{ 12}}}}}

6 0
2 years ago
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